AP EAMCET · Maths · Inverse Trigonometric Functions
If \(y=\operatorname{Tan}^{-1} \sqrt{x^2-1}+\operatorname{Sinh}^{-1} \sqrt{x^2-1}, x>1\), then \(\frac{d y}{d x}=\)
- A \(\frac{1}{x \sqrt{x^2-1}}\)
- B \(\frac{x+1}{x \sqrt{x^2-1}}\)
- C \(\frac{x+1}{x^2 \sqrt{x^2-1}}\)
- D \(\frac{\mathrm{x}}{\sqrt{\mathrm{x}^2-1}}\)
Answer & Solution
Correct Answer
(B) \(\frac{x+1}{x \sqrt{x^2-1}}\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x} = \frac{1}{1+(\sqrt{x^2-1})^2} \cdot \frac{x}{\sqrt{x^2-1}} + \frac{1}{\sqrt{1+(\sqrt{x^2-1})^2}} \cdot \frac{x}{\sqrt{x^2-1}}\) \(\frac{d y}{d x} = \frac{1}{x^2} \cdot \frac{x}{\sqrt{x^2-1}} + \frac{1}{\sqrt{x^2}} \cdot \frac{x}{\sqrt{x^2-1}}\)…
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